Question

$PbC{l}_2$ is insoluble in cold water. Addition of $HCl$ increases its solubility due

• A. Formation of soluble complex anions like $[PbC{l}_3{]}^{-}$
• B. Oxidation of $Pb(II)$ to $Pb(IV)$
• C. Formation of $[Pb(H_{2}O{)}_6{]}^{2+}$
• D. Formation of polymeric lead complexes

Answer 1

Answer: (A)

$PbC{l}_2$ react with $HCl$ as follows
$PbC{l}_2(s)+C{l}^{-}\stackrel{\text{ Cold }}{\to }{\left[PbC{l}_3\right]}^{-}(aq)$
$PbC{l}_2(s)+2C{l}^{-}\underset{\text{of HCl}}{\overset{\text{ Excess }}{\to }}{\left[PbC{l}_4\right]}^{2-}(aq)$
Thus, addition of excess of $C{l}^{-}$ions change the $PbC{l}_2$ as soluble complex of ${\left[PbC{l}_4\right]}^{-2}$.
Hence, becomes soluble.